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To find the center frequency of the auditory filter whose critical bandwidth or ERB is equal to a given value, the ERB equation (Equation 2.6), which relates the critical bandwidth to the center frequency of the filter (f_c), can be rearranged as follows.

Equation 2.6 states the following.

ERB = {24.7 3 [(4.37 3 f_c) + 1]} Hz

(A2.1)

where f_c = the filter center frequency in kHz
ERB = the equivalent rectangular bandwidth in Hz
equation valid for (100 Hz < ƒ_c < 10,000 Hz)

Rearranging step by step to find f_c as a function of ERB:

\frac{ERB}{24.7} = (4 .37 × f_{c}) + 1

$\frac{ERB}{24.7} = (4 .37 × f_{c}) + 1$

Therefore:

(\frac{ERB}{24.7}) - 1 =(4 .37 ​ × f_{c})

$(\frac{ERB}{24.7}) - 1 =(4 .37 × f_{c})$

Therefore:

[\frac{(\frac{ERB}{24.7}) - 1}{4.37}] = f_{c} (kHz)

$[\frac{(\frac{ERB}{24.7}) - 1}{4.37}] = f_{c} (kHz)$

So that:

f_{c} (k H z) = [\frac{(\frac{E R B}{24.7}) - 1}{4.37}]

$f_{c} (k H z) = [\frac{(\frac{E R B}{24.7}) - 1}{4.37}]$

(A2.2)

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Table of Contents for
Appendix 2 Solving the ERB Equation

Appendix 2: Solving the ERB Equation

Table of Contents for Appendix 2 Solving the ERB Equation

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Appendix 2: Solving the ERB Equation

Table of Contents for
Appendix 2 Solving the ERB Equation